Find the value of constant c for which line y=2x+c is a tangent to curve y^2=4x?

im doing as level math ..i have seen the solution but am unable to solve it ..it has been solved using the dicriminant


fist they eliminated the variable y and then used the discriminant to find c which is positive 1/2

Find the value of constant c for which line y=2x+c is a tangent to curve y^2=4x?
so when you see the word tangent the first thing that should come to mind is the derivative. The derivative gives all the slopes of the original graph, so, the problem is basically asking ...which point has a slope of 2....





isolate y to get y = sqrt(4x)


find it's derivative y' = 2 / sqrt(4x)





set the derivative to 2 .....2 = 2 / sqrt(4x)





x= 1/4.....so in the original function....a x=1/4 it has a slope of 2. now you need to find it's coordinates in the original function.





y = sqrt(4X) y = sqrt( 4 * 1/4 ) = sqrt 1 = 1





plug 1/4 in the original to get y= 1...this is what you get (1/4,1)





set up y = mx + c 1 = 2(1/4) + c





ANSWER :


c = 1/2
Reply:I don't know why they would do it that way.





First just isolate y in the second equation:





y=2√(x)=2x^(1/2)





You know the tangent line's equation is y=2x+c, and therefore its slope is 2. This means the function's derivative is 2.





So we find the derivative:





y=2x^(1/2)


y'=1/√(x)





Now set that equal to 2:





2=1/√(x)


2√(x)=1


√(x)=1/2


x=1/4





So we have the x-value where that given line is tangent to the curve (1/4). Plug that into the original equation to get the y-value:





y=2√(x)


y=2√(1/4)


y=1/2





That means the point of tangency is (1/4,1/2). Plug that into our equation for the tangent line to find c:





y=2x+c


1/2=2(1/4)+c


1/2=1/2+c


0=c





So I got c=0... but it's late, so I may have missed something stupidly obvious. Don't trust me until you check it :P .





-IMP ;) :)
Reply:Rewrite y^2=4x as y=(4x)^0.5 = y=2x^0.5


If the line touch the curve, they have a common point, we can find that making 2x+c=2x^0.5





Now we know that c=2x^0.5-2x (I)





The angular coeficient of the line is the derivative of the function on the point the line touch the curve, and with the expression y=2x+c we can that this coeficient is 2.


So...


derivative of y is y'=1/(x)^0.5


now solving 1/(x)^0.5=2 give us x=1/4 (II)





using (II) in (I) we have that c=1/2